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Welcome to lesson 2 on Power System Analysis.
In this lesson, we will talk about power
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system representation. Well, in power system
since it is a very complicated system. We
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use to for modeling, we represent the system
in a simpler and easier to follow forms.
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.
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.And for that we will talk about how we represent
this power system in. In this lesson 2,
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we will start with introduction to single
line diagram. Then, we will talk about the
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symbols which are used for various power system
components, in single line diagram.
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Then we will go on building single line diagram
for power system. After that we will go
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to the per unit representation of power system.
We will talk about how to select the base
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quantities. And finally, we will talk about
the advantages of a per unit system.
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.
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Well, once this you have gone through this
lesson, you would be able to explain the
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concepts of single line diagram. And what
are it is advantages. You will be able to
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build
a single line diagram for a power system.
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You will be also able to explain the concept
of
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per unit system of normalization.
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..
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And how to select the base quantities for
the per unit system, and a power system having
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different voltage levels and different power
ratings of the equipment. Then, we will also
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be able to compute per unit impedance values
for power system components, having
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different base values. And finally, you would
be able to explain the advantages of per
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unit systems of representation.
.
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So, we will take up these two forms of representation
of a power system. The first one is
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the graphical representation, in terms of
single line diagram. And the second one will
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be
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.how to represent the various electrical quantities,
in terms of per unit values. So, first we
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will talk about the single line diagram for
a power system.
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.
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Well, as we have said in lesson 1. Earlier
power systems are extremely complicated
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electrical networks. In fact, they are most
probably the most complicated man-made
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systems. Another problem that these power
system network is that, we always deal with
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3 phase networks. That is all the lines are
3 phases means, each power circuit consists
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of
three conductors. All the devices are installed
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in all the 3 phases. A complete
conventional diagram showing, all the connections
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is going to be very complicated.
And it will be impractical, because it will
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be very difficult to read this diagram. And
understand the connections between various
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components, because of the complications.
Which will be coming because of so many lines
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had to be shown on the diagram. Now,
we will try to take advantage of some of the
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basic characteristics of the power system,
in
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trying to simplify the system for representing
it.
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..
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We know that 3 phase systems are designed
as balance system. And are generally
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operated as a balanced 3 phase system, which
means that all the equipment which are the
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3 phase equipment like generators, motors,
transformers. All are designed to be balanced
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system means. All the 3 phases have equal
impedances have the same voltage levels and
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so on.
And also wherever we use even single phase
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systems, like your residential loads. There
also we try to put these loads, in such a
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way that, they are equally divided on all
the 3
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phases. And so the system is more or less
a balanced system. Now, if we have a balanced
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system. Then what we have is all the 3 phases
are having the same currents and voltages.
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And therefore, whatever we have, whatever
values of currents and voltages in one phase
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is there, is going to be in the other two
phases. Except that they will be out of phase
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by
120 degree from each other. And therefore,
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it is possible to make a single phase, or
a
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single line representation of this balanced
3 phase system. This is the property that
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we try
to take make use of in building single line
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diagrams.
Now, single line diagrams are concise way
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of communicating the basic arrangement of
power system components. That is what we need
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in case of representing graphically the
power system network is; to see the basic
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interconnections of various components. So,
that we know what is connected where and how?
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So, this is what we try to do using the
single line diagram.
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.Now, a single line diagrams which are also
called SLD’s, use single line to represent
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all
the 3 phases. As I said earlier, that 3 phases
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since they are all of same kind having same
similar currents and voltages. Therefore,
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we use a single line to represent all the
3
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phases.
.
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Since, we use only one line to represent all
the 3 phases. Single diagrams are also called
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one line diagrams. And these diagrams show
the relative electrical interconnections of
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various components. Such as generators, transformers,
transmission and distribution
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lines, loads, circuit breakers, and all these
components which go in to make the power
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system network. Now, single line diagram when
we are using the graphical
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representation, we need to use symbols for
the various power system components,
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although there is no universally accepted
set of symbols.
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..
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But, we use certain symbols which are more
or less. Universal in the sense, that most
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of
the people use these kinds of symbols. Generators
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is represented like this. Similarly,
transformers 2-winding transformers are represented
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with 2-windings, shown 3-winding
transformers. With 3-windings like this and
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auto transformers are shown in this fashion.
.
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We have symbols for current transformers,
potential transformers, disconnect, isolators,
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which is circuit breakers, fuses, reactors,
lighting arresters. All these components,
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different symbols are used for these components.
And by combining these symbols, we
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build the power system network diagram.
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..
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Now, in power system we have a concept of
a bus-bar. Bus-bars are nothing but, nodes
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in electrical circuit. So, basically in power
system, we call these nodes of electrical
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circuit as the bus-bars. This is required
in power system, because you cannot join various
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lines, which may be very thick lines just
by connecting at one point. So, what is used
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is a
long bar is used. And these various lines
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and components are connected to this bar by
means of nuts and bolts.
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This was the system which was used from very
beginning. And it is still continuing
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though if you go for go and see a high voltage
substation. You will find bus-bars are very
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similar to transmission lines. They are thick
aluminum conductors put on overhead
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towers and the various lines and other components
are joined to this. So, basically
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electrically if you see a bus-bar is nothing
but, a node.
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Now, in single line diagrams bus-bars are
used or represented as short lines, short
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thick
lines which is used perpendicular to the transmission
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lines. And is connecting these lines.
So, when we show different lines connected
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at a bus-bar. What we have is the lines
normally will be shown as the horizontal lines.
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Then the bus-bar will be a vertical line,
somewhat thicker than the transmission lines
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which are being shown.
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..
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Now, here for an example, I am showing you
a single line diagram of a power system.
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Now, in this diagram, you see we have a generator,
we have circuit breakers. Then we
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have a bus-bar which is connecting this generator
to various transformers. We have
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transformers connected to the bus-bar through
circuit breakers. We have another
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transformer here for the other line. And the
lines the two parallel lines are connected
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to
the transformers through circuit breakers
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shown here.
Then, we have another transformer on the other
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side. And this is again connected
through circuit breakers to a bus-bar to which
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a load may be connected. So, here what we
see is how we represent a power system network,
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which is consisting of a generator
connected through two step up transformers;
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connecting high voltage transmission line.
And then we have step down transformers to
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reduce the voltage, connecting to a low
voltage bus-bar to which load is connected.
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So, in this way we can build a single line
diagram. And now I will show you how we
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build a single line diagram using the symbols.
Here, we have taken an example of a
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generator 3 phase generator. Connected by
means of step, two step up transformers to
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a
double circuit transmission line. Then through
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two step down transformers to a low
voltage bus-bar to which load is connected.
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So, what we have is, we have used the generator
symbol here. Then, we have a circuit
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breaker with for protection of the generator,
connects it to a bus-bar shown here. Then,
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we have again circuit breakers to circuit
breakers on the primary side of the transformer.
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.Or the low voltage side for the two transformers.
And then on the secondary side or the
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high voltage side again this is transformer
is connected, by means for these circuit
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breakers to the two high voltage transmission
lines, which again at the other end, through
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the circuit breakers is connected to the high
voltage side of the step down transformers.
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And the low voltage side of the step down
transformers is connected to a bus-bar by
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means of these two circuit breakers. And on
this low voltage bus, we have the load.
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.
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Next we have a substation with 2 bus-bars
and 4 transmission lines. We have these
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transmission line 1, 2, 3 and 4 connected
to these two bus-bars. By means of circuit
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breakers shown here, and the isolator switches
shown here. So, this is how we represent
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a power system network, using single line
diagram. This is much simpler, much easier
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to
understand, because it is showing you the
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relative interconnection of the various
components. And it is much more simplified
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and easy to see and draw.
Next, we will talk about representing the
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power system components with it is electrical
quantities. Most of the time, we represent
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these electrical quantities, in terms of power
system per unit system. One of the reason
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for this is, because power system consists
of
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equipments which are very wide varying ratings,
and if we use all these different voltage
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levels and the power levels. Then, there is
always a chance of making mistakes, because
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we are talking in terms of somewhere kilowatts,
somewhere megawatts, somewhere
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kilovolts, somewhere just volts. So, chances
of making mistakes is much higher. But, if
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.we represent these values in terms of per
unit, we will see what advantages we get out
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of
this.
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.
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So, in power system electrical quantities
such as power, voltage, current, impedance
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etcetera, are very often expressed as per
unit of a base or reference value. That is
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what
we do is we represent these values on a reference,
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with respect to a reference or a base
value, per unit value is expressed by the
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equation. Per unit quantity is equal to actual
quantity divided by the base value of that
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quantity.
.
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.Now, we have for interest 4 values. That
is power, voltage, current and impedance,
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the
out of these 4 values, if we choose any two.
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The other two can always be represented in
terms of these two. That is we need to choose
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only 2 values independently, the other 2
values will come out to be dependent on these.
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The two independent base values which
are selected for per unit representation in
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power system, is generally a base voltage
value
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and the complex power base value.
So, we use V base and S base as our reference
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values. And the values for Z base, that is
the impedance base. And the current base I
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base will get derived from V base and S base.
Now, let us start with a single phase system.
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For a single phase system we have P base
single phase is equal to Q base single phase
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is equal to S base single phase. That means,
we choose the complex power base. And that
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base is going to be that value is also the
base value for the real, as well as reactive
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power. Now, the base value for the current,
I
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base will come out to be S base by V base.
That is S base single phase by V base lying
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to
neutral or a single phase value.
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.
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The impedance base value Z base; Z b is equal
to R b the resistance base or the reactance
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base. All of them will have the same base
value. And this is equal to V base lying to
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neutral by I base. This can also be written
as V base lying to neutral square divided
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by S
base single phase. Of course, if we are talking
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about admittance, admittance will be just
the inverse of the impedance. So, Y base is
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equal to G base the conductance base or the
susceptance base B base. And that will be
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equal to 1 by Z base.
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.This is very simple. But, in power system
what we have is, we use large number of
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transformers for voltage transformation, from
low voltage to high voltage, and then from
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high voltage to low voltage at various points.
Therefore, what we need to do is, we need
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to choose these base values very properly.
One of the convention that we follow for
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choosing the base values, for per unit system
is that value of S base S b is same for the
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entire system. That is the base value for
the complex power is same for the whole
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system.
Now, the ratio of the base value for the voltage.
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On either side of a transformer is
selected to be same, as the ratio of the transformer
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voltage ratings. This is very
important, because on the two sides of the
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transformer you have two different voltage
levels. And we can choose different voltage
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bases on the two sides. It is very
advantageous; if we use these bases on the
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two sides of the transformer. The voltage
bases on the two sides of the transformer
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to be in the same ratio, as the turns ratio
of the
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transformer. Generally, what we try to do
is, we choose the base voltage values on the
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two sides of the transformer, as the nominal
or the rated voltage of the transformer on
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the
two sides. This we will see later has lots
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of advantages.
.
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Now, let us see we have a transformer. Now,
this transformer can be represented as a
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leakage impedance. That is R in per unit and
j in per unit referred to the primary side.
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And an ideal transformer and then we have
the voltage V 2 per unit. On the secondary
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side and I 2 per unit current on the secondary
side. We have neglected the magnetizing
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.part of the transformer here. Most of the
time in power system analysis, unless we are
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interested in finding out the losses in the
core losses as such.
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We most of the time will be neglecting that.
The core loss or the magnetizing part of the
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transformer winding. Because, it is effect
is very, very small and neglecting this does
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not
cause much errors. And also it simplifies
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our diagram considerably. Now, if you look
at
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this circuit diagram. Then we can write, V
1 per unit is equal to V 1 by V 1 base, which
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is equal to N 1 by N 2 in to V 2. Because
V 1 is equal to N 1 by N 2 in to V 2 and this
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divided by V 1 base.
Now, if we are choosing our two base values,
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on the two sides of the transformers base
voltage values. V 1 base divided by V 2 base
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is equal to V rated 1 divided by V rated 2,
this is equal to N 1 by N 2. This is what
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as I said if we follow these conventions.
That
209
00:24:38,490 --> 00:24:45,080
base values for the voltage at the two sides
of the transformer is in the ratio of the
210
00:24:45,080 --> 00:24:49,250
rated
voltages. Which will be also the turns ratio
211
00:24:49,250 --> 00:24:50,960
of the transformer.
.
212
00:24:50,960 --> 00:25:01,290
Then, we have V 1 per unit is equal to N 1
by N 2 in to V 2 divided by V 1 base. Now,
213
00:25:01,290 --> 00:25:04,840
V
1 base will become N 1 by N 2 in to V 2 base.
214
00:25:04,840 --> 00:25:13,420
Because, V 1 base by V 2 base is equal to
N 1 by N 2. Therefore, this is equal to V
215
00:25:13,420 --> 00:25:23,530
2 by V 2 base which is equal to V 2 per unit.
That is what we find is the voltage at the
216
00:25:23,530 --> 00:25:31,590
two sides of the transformer in a per unit
values, are going to be same, though the actual
217
00:25:31,590 --> 00:25:39,559
values of the voltages will be quite
different. Because, one side for a 11 KV to
218
00:25:39,559 --> 00:25:44,700
132 KV transformer. The low voltage side
value will be somewhere around 11 KV whereas,
219
00:25:44,700 --> 00:25:47,640
the high voltage side value will be
220
00:25:47,640 --> 00:25:54,450
.around 132 KV. But, in case of per unit system
if we have chosen the base value of
221
00:25:54,450 --> 00:26:02,710
voltage, on the low voltage side as 11 KV
and on 132 KV side. As 132 KV, then on both
222
00:26:02,710 --> 00:26:07,220
sides we have the value of the voltage equal
to 1 per unit.
223
00:26:07,220 --> 00:26:17,890
So, this in per unit system, the transformer
voltages are going to be same. Similarly,
224
00:26:17,890 --> 00:26:21,559
we
can write I 1 per unit is equal to I 1 by
225
00:26:21,559 --> 00:26:26,500
I 1 base, which is equal to N 2 by N 1 into
I 2 that
226
00:26:26,500 --> 00:26:36,330
is I 1 divided by I 1 base. Now, using I 1
base is equal to S base by V 1 base. This
227
00:26:36,330 --> 00:26:40,090
will be
equal to S base divided by N 1 by N 2 into
228
00:26:40,090 --> 00:26:54,570
V 2 base, this finally comes to N 2 by N 1
into I 2 base. That is we are seeing that
229
00:26:54,570 --> 00:27:03,650
I 1 base and I 2 base are related by a relationship
I 1 base is equal to N 2 by N 1 into I 2 base.
230
00:27:03,650 --> 00:27:08,251
That is it is in the inverse ratio of the
base
231
00:27:08,251 --> 00:27:16,370
values. That we had chosen for the voltage.
Where we had V 1 base by V 2 base is equal
232
00:27:16,370 --> 00:27:24,620
to N 1 by N 2. Here we have I 1 base by I
2 base is equal to N 2 by N 1.
233
00:27:24,620 --> 00:27:25,620
.
234
00:27:25,620 --> 00:27:32,550
Now, if you look at the per unit quantities
I can write I 1 per unit is equal to N 2 by
235
00:27:32,550 --> 00:27:36,919
N 1
in to I 2 divided by I 1 base. Now, I 1 base
236
00:27:36,919 --> 00:27:45,010
is equal to N 2 by N 1 into I 2 base.
Therefore, it comes out to b I 2 by I 2 base
237
00:27:45,010 --> 00:27:48,419
which is nothing but, I 2 per unit. That is
the
238
00:27:48,419 --> 00:27:58,190
current in per unit values for I 1 and I 2
is going to be same, which shows since the
239
00:27:58,190 --> 00:28:04,470
voltage in per unit is also same. And the
current is also same which shows this ideal
240
00:28:04,470 --> 00:28:14,240
transformer is simply a 1 is to 1 transformer
in per unit system. So, it can as well be
241
00:28:14,240 --> 00:28:21,039
neglected.
So, what we find that, if we have used a per
242
00:28:21,039 --> 00:28:27,170
unit system with proper voltage base
selection. Then this auto transformer or the
243
00:28:27,170 --> 00:28:28,720
turns transformation, the voltage
244
00:28:28,720 --> 00:28:36,270
.transformation ratio, on current transformation
ratio which comes for the transformers
245
00:28:36,270 --> 00:28:42,560
are no longer required in per unit system.
So, transformers can be simply eliminated.
246
00:28:42,560 --> 00:28:52,000
And it is leakage impedance can be put in
per unit values. Now, the same thing we can
247
00:28:52,000 --> 00:28:58,049
see for the impedance. The transformer impedance
on the primary referred to the primary
248
00:28:58,049 --> 00:29:10,180
side is Z 1 per unit will be equal to Z 1
into I 1 base divided by V 1 base.
249
00:29:10,180 --> 00:29:11,180
.
250
00:29:11,180 --> 00:29:22,100
And if it is referred to the other side, secondary
side, then Z 2 is equal to Z 1 in to N 2 by
251
00:29:22,100 --> 00:29:31,830
N 1 square. Therefore, Z 2 per unit will be
equal to Z 1 into N 2 by N 1 square which
252
00:29:31,830 --> 00:29:37,460
is
nothing but, Z 2 into I 2 base by V 2 base,
253
00:29:37,460 --> 00:29:41,700
this well again substituting the values we
have
254
00:29:41,700 --> 00:29:49,950
Z 1 in to N 2 by N 1 square into I 1 base
into N 1 by N 2. For I 2 base we can write
255
00:29:49,950 --> 00:29:53,669
and
divide by V 2 base can be written as V 1 base
256
00:29:53,669 --> 00:30:00,220
into N 2 by N 1.
Which finally, comes out to be Z 1 into I
257
00:30:00,220 --> 00:30:03,800
1 base divided by V 1 base which is equal
to Z
258
00:30:03,800 --> 00:30:11,780
1 per unit. That is what we are seeing is
that, even the impedance referred to on the
259
00:30:11,780 --> 00:30:16,210
two
sides in per unit system are going to be same.
260
00:30:16,210 --> 00:30:22,930
And therefore, the transformation ratios for
referring the impedance from one side to the
261
00:30:22,930 --> 00:30:29,660
other side of the transformer is no longer
required. So, whether the impedance is referred
262
00:30:29,660 --> 00:30:35,260
to on the primary side. Or on the
secondary side of a transformer, it does not
263
00:30:35,260 --> 00:30:40,419
make any difference in per unit system,
because it is going to be same on both side.
264
00:30:40,419 --> 00:30:48,170
And that is why we can very easily eliminate
all these transformation ratios, which we
265
00:30:48,170 --> 00:30:53,520
have to do in actual system. This is a very
great
266
00:30:53,520 --> 00:31:00,110
advantage of per unit system of representation,
because in a power network we have very
267
00:31:00,110 --> 00:31:05,260
.large number of transformers. And we have,
if we are working with actual quantities,
268
00:31:05,260 --> 00:31:07,910
we
will have to take care of all these turns
269
00:31:07,910 --> 00:31:13,310
ratios all the time.
.
270
00:31:13,310 --> 00:31:20,970
In power system, we also use equipments at
various with various ratings, different
271
00:31:20,970 --> 00:31:28,480
ratings. And therefore, we need to convert
the impedance values from one rating to
272
00:31:28,480 --> 00:31:36,809
another. Because as I said earlier the equipment
manufacturers will generally specify the
273
00:31:36,809 --> 00:31:42,840
value of the impedance of any equipment, in
terms of percentage, or per unit values
274
00:31:42,840 --> 00:31:53,060
based on the name plate rating of that equipment.
We might choose a different base for power,
275
00:31:53,060 --> 00:32:00,419
for the whole power system. We may have
a difference base value for the voltage of
276
00:32:00,419 --> 00:32:04,990
the system in that part. Therefore, we need
to
277
00:32:04,990 --> 00:32:13,760
convert the impedance from one base value
to another base value. So, suppose we have
278
00:32:13,760 --> 00:32:22,600
the value of impedance given at one base,
then how do we convert it to another base.
279
00:32:22,600 --> 00:32:27,530
So,
we have this relationship Z per unit new value
280
00:32:27,530 --> 00:32:38,049
is equal to Z actual value by Z base new,
which is equal to Z per unit old into Z base
281
00:32:38,049 --> 00:32:41,480
old which is the actual value. Because, if
we
282
00:32:41,480 --> 00:32:47,679
multiply the per unit value by the base value,
we get the actual value. So, this is Z actual
283
00:32:47,679 --> 00:32:56,419
is Z per unit old in to Z base of old divided
by Z base new. This can be written as Z per
284
00:32:56,419 --> 00:33:08,820
unit old into V base old by V base new square
into S base new by S base old. When we
285
00:33:08,820 --> 00:33:22,059
substitute for Z base old as V base old square
by S base and Z base new as V base new
286
00:33:22,059 --> 00:33:28,340
square by S base new.
If we substitute that, this is the relationship
287
00:33:28,340 --> 00:33:37,430
that we will get Z base Z per unit new is
equal to Z per unit old into V base old divided
288
00:33:37,430 --> 00:33:41,380
by V base new square into Z base new by
289
00:33:41,380 --> 00:33:49,710
.Z base old. So, if we know the old values
or the equipment value at particular base.
290
00:33:49,710 --> 00:33:53,789
And
we want to convert it to other base, we can
291
00:33:53,789 --> 00:33:59,299
find out the value of the impedance at this
new base values.
292
00:33:59,299 --> 00:34:00,299
.
293
00:34:00,299 --> 00:34:09,010
Till now we talked about single phase system.
The per unit system representation can be
294
00:34:09,010 --> 00:34:17,210
extended. Or used in 3 phase system also in
a similar way. What we have is for a single
295
00:34:17,210 --> 00:34:23,809
phase system, if we have chosen a base S base
for single phase, this is equal to S base
296
00:34:23,809 --> 00:34:27,960
for
3 phase system divided by 3. We know that
297
00:34:27,960 --> 00:34:34,819
for any equipment rating. If we have the 3
phase rating, then per phase rating is going
298
00:34:34,819 --> 00:34:42,569
to be 1 3rd of that.
So, similarly we can choose the 3 phase base
299
00:34:42,569 --> 00:34:49,909
for complex power, which will again be
valid or will be same for the real power,
300
00:34:49,909 --> 00:34:53,240
as well as the reactive power. So, S base
3 phase
301
00:34:53,240 --> 00:35:01,150
is equal to P base 3 phase is equal to Q base
3 phase. In case of 3 phase system instead
302
00:35:01,150 --> 00:35:04,789
of
using line to neutral voltage, which is the
303
00:35:04,789 --> 00:35:09,670
per phase voltage in the system. We use line
to
304
00:35:09,670 --> 00:35:15,130
line voltage and the relationship is V base
line to neutral is equal to V base line to
305
00:35:15,130 --> 00:35:19,479
line
divided by root 3. Or V base line to line
306
00:35:19,479 --> 00:35:29,289
is equal to root 3 times V base line to neutral.
Therefore, we get I base which is equal to
307
00:35:29,289 --> 00:35:32,960
S base single phase divided by V base line
to
308
00:35:32,960 --> 00:35:42,450
neutral as equal to S base 3 phase divided
by root 3 V base line to line. So, if we have
309
00:35:42,450 --> 00:35:46,640
I
base known, then we can write Z base is equal
310
00:35:46,640 --> 00:35:53,089
to V base line to neutral by I base, which
is same as V base line to neutral square divided
311
00:35:53,089 --> 00:36:01,609
by S base single phase. And if we are
using 3 phase system, then this multiplied
312
00:36:01,609 --> 00:36:07,849
by 3 and this multiplied by 3 will give me
V
313
00:36:07,849 --> 00:36:09,700
base line to line square.
314
00:36:09,700 --> 00:36:16,839
.Because, V base line to line is root 3 times
V base line to neutral divided by S base 3
315
00:36:16,839 --> 00:36:24,999
phase. Therefore, Z base for a single phase
system, as well as 3 phase system is going
316
00:36:24,999 --> 00:36:29,960
to
be same. Except that the bases for 3 phase
317
00:36:29,960 --> 00:36:36,829
system has to be chosen as S base for 3 phase
is three times the S base for single phase.
318
00:36:36,829 --> 00:36:48,979
And V base for 3 phase system should be the
line to line voltage phase. Again Z base is
319
00:36:48,979 --> 00:36:54,619
same as the base for resistance, reactance
or
320
00:36:54,619 --> 00:37:01,470
the impedance. And if we take the reciprocal
of it, we get the base value for the
321
00:37:01,470 --> 00:37:10,150
admittance.
So, we can use this per unit system both in
322
00:37:10,150 --> 00:37:16,469
single phase. Or 3 phase system normally
power systems are 3 phase system. So, we will
323
00:37:16,469 --> 00:37:24,569
be using the 3 phase system, where we
will be using the power base as 3 phase power
324
00:37:24,569 --> 00:37:32,249
base. And we will be using the voltage
base as the line to line voltage.
325
00:37:32,249 --> 00:37:33,249
.
326
00:37:33,249 --> 00:37:44,240
Now, let us take a simple example for this
per unit representation. Here we have taken
327
00:37:44,240 --> 00:37:48,150
a
problem where we say that prepare, a per phase
328
00:37:48,150 --> 00:37:53,969
schematic of the system shown in the
figure, which is shown in the next slide.
329
00:37:53,969 --> 00:38:03,819
On the next slide, and show all the impedances
in per unit on a 100 MVA base. 132 KV voltage
330
00:38:03,819 --> 00:38:12,229
base in the transmission line of the
circuit. That is the impedance the power base
331
00:38:12,229 --> 00:38:16,460
for the whole system is chosen as 100
MVA.
332
00:38:16,460 --> 00:38:24,630
Whereas, the voltage base in the transmission
line circuit is chosen as 132 KV. The
333
00:38:24,630 --> 00:38:34,039
voltage base in the other parts of the circuit,
as we will see will depend on the voltage
334
00:38:34,039 --> 00:38:39,380
transformation ratio on the two sides of the
transformers. Now, for this system we are
335
00:38:39,380 --> 00:38:49,099
.given that we have a generator G 1 with 100
MVA rating at a 11 KV and impedance of
336
00:38:49,099 --> 00:38:59,349
0.15 per unit, another generator G 2 with
a 200 MVA rating. And voltage rating of 13.8
337
00:38:59,349 --> 00:39:09,660
KV and reactance X is equal to 0.2 per unit.
We have a transformer T 1 with a rating of
338
00:39:09,660 --> 00:39:18,799
120 MVA. And voltage ratio of 1 KV to 132
KV and the reactance of 0.1 per unit, another
339
00:39:18,799 --> 00:39:27,329
transformer T 2 with a power rating of 250
MVA. And a voltage transformation ratio of
340
00:39:27,329 --> 00:39:37,769
13.8 KV to 132 KV. And a reactance of 0.1
per unit on its base, that is on it is rated
341
00:39:37,769 --> 00:39:45,759
value base. We have a load connected to the
system which has a value of 250 MVA at 0.8
342
00:39:45,759 --> 00:39:50,900
per factor lagging, and operating at 132
KV.
343
00:39:50,900 --> 00:39:51,900
.
344
00:39:51,900 --> 00:39:58,329
So, this is our system we have a generator
G 1, connected through a transformer T 1 to
345
00:39:58,329 --> 00:40:04,380
this bus. And a generator G 2, connected through
transformer T 2 through this bus. This
346
00:40:04,380 --> 00:40:11,059
is, these two transformers are step up transformers.
And we have three transmission
347
00:40:11,059 --> 00:40:19,920
lines, working at 132 KV and a load of 200
MVA at 0.8 power factor lagging connected
348
00:40:19,920 --> 00:40:28,609
at this bus. Now, let us see how we convert
this in per unit system. The impedance
349
00:40:28,609 --> 00:40:34,950
values, here for the transmission lines are
given in ohms.
350
00:40:34,950 --> 00:40:35,950
..
351
00:40:35,950 --> 00:40:41,410
So, what we have to do is determine the per
unit impedance of the load for the following
352
00:40:41,410 --> 00:40:48,739
cases. Load model has series combination of
resistance and inductance. That is we can
353
00:40:48,739 --> 00:40:56,549
put this load as a series combination. Or
we can put this load as a parallel combination
354
00:40:56,549 --> 00:40:58,560
of
resistance and inductance.
355
00:40:58,560 --> 00:40:59,560
.
356
00:40:59,560 --> 00:41:08,459
So, what we do in solving this problem. First
all the data represented are in per unit on
357
00:41:08,459 --> 00:41:12,410
a
base specified for each component. So, the
358
00:41:12,410 --> 00:41:20,249
base values for the reactance’s are generators
transformers, all of them are given on their
359
00:41:20,249 --> 00:41:24,239
own rating. So, in our analysis we will
360
00:41:24,239 --> 00:41:29,510
.convert all these quantities to a common
system base specified in the problem, which
361
00:41:29,510 --> 00:41:35,400
is
100 MVA. Now, base KV in the transmission
362
00:41:35,400 --> 00:41:40,680
line part is 132 KV.
Base KV in the generator circuit G 1 will
363
00:41:40,680 --> 00:41:49,780
be 132 in to 11 by 132, which is the voltage
transformation ratio for the transformer.
364
00:41:49,780 --> 00:41:55,710
So, it will be 11 KV on the generator side.
Similarly, base KV in the generator circuit
365
00:41:55,710 --> 00:42:09,459
G 2 is going to be 132 into 13.8divided by
161, this is equal to 11.31 KV. So, base values
366
00:42:09,459 --> 00:42:20,989
on two sides of the transformers are to be
chosen based on the voltage ratio of the transformer
367
00:42:20,989 --> 00:42:28,450
or the nominal ratio of the
transformer or the turns ratio of the transformer.
368
00:42:28,450 --> 00:42:29,450
.
369
00:42:29,450 --> 00:42:35,920
Now, we convert all the parameter values to
per unit on the common base specified. So,
370
00:42:35,920 --> 00:42:44,839
G 1 the value given is X is equal to 0.15.
Now, we are converting it to our new base.
371
00:42:44,839 --> 00:42:50,309
So,
S base old by S base new is both are 100.
372
00:42:50,309 --> 00:42:58,920
Similarly, the voltage also base old and base
new voltage are also same. So, the value of
373
00:42:58,920 --> 00:43:04,819
the impedance remains same. So, using the
same relationship, we find out the value of
374
00:43:04,819 --> 00:43:10,789
reactance for G 2 which comes out to be
0.1489 instead of 0.2
375
00:43:10,789 --> 00:43:19,469
Similarly, when we convert it for the transformers
we get 41 the value on 100 MVA
376
00:43:19,469 --> 00:43:31,969
base, as 0.0833 per unit. And for T 2 we get
0.05955 per unit. The base impedance of the
377
00:43:31,969 --> 00:43:39,819
transmission line circuit is going to be 132
KV square by 100, because the voltage base
378
00:43:39,819 --> 00:43:46,230
is 132 KV for the transmission line. And the
MVA base for the whole system is chosen
379
00:43:46,230 --> 00:43:53,140
as 100. Therefore, the base impedance is KV
square, base KV square divided by base
380
00:43:53,140 --> 00:44:02,519
.MVA, which is equal to 174.24 ohms. So, we
know the Z base for the transmission line
381
00:44:02,519 --> 00:44:05,599
circuit as 174.24 ohms.
.
382
00:44:05,599 --> 00:44:17,309
Therefore, we can find out for each transmission
line, the per unit value of the
383
00:44:17,309 --> 00:44:27,039
impedance. So, for the transmission line having
impedance 50 plus j 200, the per unit
384
00:44:27,039 --> 00:44:34,609
value comes out to be 0.2870 plus j 1.1478
per unit. Similarly, for the transmission
385
00:44:34,609 --> 00:44:38,589
line
having impedance value of 25 plus j 10, the
386
00:44:38,589 --> 00:44:48,880
value comes out to be 0.1435 plus j 0.5739
per unit.
387
00:44:48,880 --> 00:44:49,880
.
388
00:44:49,880 --> 00:44:58,160
.Also we have the load is specified as 250
MVA at 0.8 per factor lagging. So, that can
389
00:44:58,160 --> 00:45:04,859
be
converted in to 200 megawatt plus j 150 means,
390
00:45:04,859 --> 00:45:11,960
150 megawatts. We now look for
different types of combinations of load resistance
391
00:45:11,960 --> 00:45:17,499
and reactance. That is we are asked to
find out the series combination as well shunt
392
00:45:17,499 --> 00:45:21,000
combination, for the impedance of this
load.
393
00:45:21,000 --> 00:45:22,000
.
394
00:45:22,000 --> 00:45:27,529
So, for series combination what we have is,
we find out the impedance we know that S is
395
00:45:27,529 --> 00:45:35,140
equal to V I conjugate. So, I conjugate if
we take, therefore I conjugate will give you
396
00:45:35,140 --> 00:45:41,499
V
by S. And therefore, if we take V square by
397
00:45:41,499 --> 00:45:48,940
S we get Z conjugate. Z load conjugate
comes out to be, because 132 KV is the voltage
398
00:45:48,940 --> 00:45:59,009
level at that. And this is the load,
therefore the load impedance turns out to
399
00:45:59,009 --> 00:46:06,599
be 55.7568 minus j 41.8176. This is Z load
conjugate the star is indicating, this is
400
00:46:06,599 --> 00:46:09,680
conjugate value.
Therefore, the actual value of the load is
401
00:46:09,680 --> 00:46:20,799
going to be 55.7568 plus j 41.8176 ohms. This
is going to be there, because this has to
402
00:46:20,799 --> 00:46:25,320
be a reactive load or a inductive load. So,
this
403
00:46:25,320 --> 00:46:31,539
plus must be there, this cannot be a negative,
because we have load which is having a
404
00:46:31,539 --> 00:46:39,499
lagging power factor. So, Z load per unit
will again we just dividing it by Z base comes
405
00:46:39,499 --> 00:46:46,239
out to be 0.32 plus j 0.24 per unit.
406
00:46:46,239 --> 00:46:47,239
..
407
00:46:47,239 --> 00:46:52,430
If we take the parallel combination, again
we can find out the value of resistance part,
408
00:46:52,430 --> 00:47:00,719
which is KV square divided by P. That comes
out to be 87.12 ohms which in per unit
409
00:47:00,719 --> 00:47:06,270
comes out to be 0.5 per unit. The reactive
part can also be found in the same way V
410
00:47:06,270 --> 00:47:15,279
square by Q, which is coming out to be 116.16
ohms which comes out to be 0.66 unit.
411
00:47:15,279 --> 00:47:16,279
.
412
00:47:16,279 --> 00:47:21,809
And therefore, if we put all these impedance
values on the single line diagram. Then we
413
00:47:21,809 --> 00:47:31,359
get this network diagram. The generator with
its impedance at j is equal to 0.15 generator
414
00:47:31,359 --> 00:47:36,660
G 1. Then we have the transformer impedance,
then we have the transmission line
415
00:47:36,660 --> 00:47:41,220
.impedance. This is the other transmission
line, this is the other transmission line.
416
00:47:41,220 --> 00:47:43,960
And
this is the transformer impedance, this is
417
00:47:43,960 --> 00:47:51,630
the generator impedance.
All these are in per unit on 100 MVA base,
418
00:47:51,630 --> 00:47:56,719
and the voltage bases are respective bases
on
419
00:47:56,719 --> 00:48:04,229
the two sides of the transform. So, using
this all these values, this transformers are
420
00:48:04,229 --> 00:48:07,019
no
longer existing in the system except for the
421
00:48:07,019 --> 00:48:14,509
leakage impedance part. The load here is
shown as the parallel combination, similarly
422
00:48:14,509 --> 00:48:19,339
a series combination can also be shown.
.
423
00:48:19,339 --> 00:48:28,229
Now, we will see what are the advantages?
After we have solved this problem. We will
424
00:48:28,229 --> 00:48:33,969
can see what are the advantages of this per
unit system? One is manufacturers specify
425
00:48:33,969 --> 00:48:36,460
the
impedance of equipment, in percent or per
426
00:48:36,460 --> 00:48:41,360
unit on the base of the nameplate rating.
That
427
00:48:41,360 --> 00:48:50,809
is what we were given is for a generator at
100 MVA base and it is rated KV the
428
00:48:50,809 --> 00:48:58,170
reactance was given. Similarly, for generator
G 2 at 200 MVA and it is rated voltage the
429
00:48:58,170 --> 00:49:03,809
reactance was given.
We had to convert them to a single base for
430
00:49:03,809 --> 00:49:14,630
our system. But, most of the manufacturers
will provide you the impedance of the equipment
431
00:49:14,630 --> 00:49:20,619
in per unit, or percent value. If you
multiply per unit by hundred you get the percent
432
00:49:20,619 --> 00:49:26,799
value. So, this is one thing which we
get. So, we know the manufacturers will provide
433
00:49:26,799 --> 00:49:34,430
in this form. This per unit impedance of
the same type of equipment with widely different
434
00:49:34,430 --> 00:49:43,479
rating, lie in a narrow range.
That means, for generators, if you see for
435
00:49:43,479 --> 00:49:49,569
a thermal generator. The value of the
impedance or the synchronous impedance will
436
00:49:49,569 --> 00:50:00,369
be of the order of 0.8, or 0.9 per unit. For
most of the generators, whatever may be their
437
00:50:00,369 --> 00:50:02,999
rating, it may be a 10 megawatt generator,
438
00:50:02,999 --> 00:50:10,130
.it may be a 500 megawatt generator. But,
the value is going to lie between say 0.5
439
00:50:10,130 --> 00:50:14,549
to 1
per unit or 1.2 per unit in that small range,
440
00:50:14,549 --> 00:50:20,290
whereas the actual value will vary
considerably from a few ohms to milliohms.
441
00:50:20,290 --> 00:50:28,749
So, this is a great advantage that even if
we do not know the exact value while designing
442
00:50:28,749 --> 00:50:37,119
a system. We can choose an average value of
per unit impedance, for the given type of
443
00:50:37,119 --> 00:50:47,200
equipment. And do all our design calculations
based on that, before we really procure the
444
00:50:47,200 --> 00:50:53,459
equipment. Another great advantage of per
unit system is, if we choose the voltage bases
445
00:50:53,459 --> 00:51:00,089
on two sides of the transformers as the turns
ratio of the transformer or the voltage ratio
446
00:51:00,089 --> 00:51:09,869
of the transformers on the two sides. Then
the transformation ratio is no longer there,
447
00:51:09,869 --> 00:51:12,819
that
is the transformer except for its leakage
448
00:51:12,819 --> 00:51:19,279
impedance does not exist.
That is transformers or it is voltage transformation
449
00:51:19,279 --> 00:51:24,700
or the impedance transformation of
the transformer is completely eliminated,
450
00:51:24,700 --> 00:51:31,819
when we work with per unit system. And this
greatly enhances the computational simplicity
451
00:51:31,819 --> 00:51:41,690
for large networks. That is our
computational effort reduces consequently.
452
00:51:41,690 --> 00:51:48,260
Anyway when you work with this per unit
system and you get all the values in per unit.
453
00:51:48,260 --> 00:51:54,430
Finally, you need to use values which are
the actual values.
454
00:51:54,430 --> 00:52:00,589
And therefore, what you need to do is, the
final values after doing all the calculations.
455
00:52:00,589 --> 00:52:06,469
And getting all the values of voltages currents,
impedances in per unit, you will finally,
456
00:52:06,469 --> 00:52:12,529
have to convert them to actual values by multiplying
them, by the respective base lines.
457
00:52:12,529 --> 00:52:20,880
So, once you do that you will get the final
values. So, when we are doing this analysis
458
00:52:20,880 --> 00:52:23,299
for
power system, what we do is? First convert
459
00:52:23,299 --> 00:52:30,680
the actual values to the per unit values.
And then do all the calculations on per unit
460
00:52:30,680 --> 00:52:39,190
values. And final results, we again convert
them to actual values to see the actual values
461
00:52:39,190 --> 00:52:46,410
of currents and voltages and then
impedances. So, this is how we work for power
462
00:52:46,410 --> 00:53:05,289
system analysis using the per unit
system.
463
00:53:05,289 --> 00:53:11,579
Thank you.
464
00:53:11,579 --> 00:53:12,579
..
465
00:53:12,579 --> 00:53:13,579
.
466
00:53:13,579 --> 00:53:19,699
Welcome to lecture 3 in Power System Analysis.
In this lesson, we will talk about
467
00:53:19,699 --> 00:53:22,829
transmission line parameters.
468
00:53:22,829 --> 00:53:23,829
..
469
00:53:23,829 --> 00:53:33,499
Before I go in to the transmission line parameters
itself. We will first talk about the
470
00:53:33,499 --> 00:53:40,130
questions that I asked at the end of lesson
2. Well, the first question was what is a
471
00:53:40,130 --> 00:53:43,329
single
line diagram? Well, the answer to that question
472
00:53:43,329 --> 00:53:49,729
is single line diagrams are concise way
for communicating, the basic arrangement of
473
00:53:49,729 --> 00:53:57,009
power system components. Single line
diagrams use a single line to represent all
474
00:53:57,009 --> 00:54:05,339
the 3 phases. And they show the relative
electrical connections of various electrical
475
00:54:05,339 --> 00:54:11,890
components, which are used in assembling the
power system.
476
00:54:11,890 --> 00:54:12,890
.
477
00:54:12,890 --> 00:54:21,329
.The second question was, how base values
of voltages are chosen on two sides of a
478
00:54:21,329 --> 00:54:27,790
transformer? Well, the answer to this question
is, ratio of base values on either side of
479
00:54:27,790 --> 00:54:32,130
a
transformer is selected to be the same as
480
00:54:32,130 --> 00:54:38,469
the ratio of the transformer voltage ratings.
Well, this is necessary because, when we use
481
00:54:38,469 --> 00:54:49,880
this voltage bases on the two sides as the
voltage ratio of the transformer. Then, we
482
00:54:49,880 --> 00:54:58,009
find that the per unit impedance on the two
sides of the transformer are same. And therefore,
483
00:54:58,009 --> 00:55:07,759
it eliminates the use of the ideal
transformer in the power systems. This helps
484
00:55:07,759 --> 00:55:17,609
considerably in reducing the calculations
for the power system circuits.
485
00:55:17,609 --> 00:55:18,609
.
486
00:55:18,609 --> 00:55:26,380
And the third question was for a 500 megawatt
22 KV generator, the per unit impedance
487
00:55:26,380 --> 00:55:33,609
is 0.8 per unit on it is own base. What is
it is per unit impedance at 100 megawatt and
488
00:55:33,609 --> 00:55:39,309
33
KV base? Well, as we had seen in lesson 2,
489
00:55:39,309 --> 00:55:45,299
the Z per unit on the new base is given as
Z
490
00:55:45,299 --> 00:55:55,160
per unit under old base multiplied by old
V base divided by new V base square
491
00:55:55,160 --> 00:56:04,869
multiplied by new MVA base divided by old
MVA base. So, using this relationship, once
492
00:56:04,869 --> 00:56:14,630
we substitute these values we will get Z per
unit at the new base of 100 megawatt. And
493
00:56:14,630 --> 00:56:24,170
33 KV as 0.8 into 22 by 33 whole square multiplied
by 100 by 500, which will come out
494
00:56:24,170 --> 00:56:28,189
to be equal to 0.0711 per unit.
495
00:56:28,189 --> 00:56:29,189
..
496
00:56:29,189 --> 00:56:38,970
Next we will talk about the electrical parameters.
Now, we are using conductors,
497
00:56:38,970 --> 00:56:46,510
overhead conductors. And these conductors
has certain amount of resistance to the
498
00:56:46,510 --> 00:56:53,599
current flowing through them. So, depending
on this resistance, there is going to be
499
00:56:53,599 --> 00:56:59,999
power loss in these conductors. Therefore,
we must find out what is the resistance of
500
00:56:59,999 --> 00:57:00,999
the
conductor
501
00:57:00,999 --> 00:57:08,759
Then, we will also talk about inductance this
comes. Because, once the current flows in
502
00:57:08,759 --> 00:57:14,589
the conductor it sets up magnetic field because
of which there is certain amount of
503
00:57:14,589 --> 00:57:21,279
inductance. Since, these conductors are at
high voltage. There is a voltage difference
504
00:57:21,279 --> 00:57:26,189
between the two phase conductors. As well
between phase conductors and ground.
505
00:57:26,189 --> 00:57:32,309
Therefore, there is capacitance involved between
them. So, we will also talk about
506
00:57:32,309 --> 00:57:40,839
capacitance. So, these three parameters are
the most important parameters for a
507
00:57:40,839 --> 00:57:45,390
transmission line, electrical modeling.
508
00:57:45,390 --> 00:57:46,390
..
509
00:57:46,390 --> 00:57:53,869
Well, as far as resistance is concerned, the
dc resistance of a conductor at any
510
00:57:53,869 --> 00:58:03,289
temperature is given by R dc is equal to rho
l by A, where rho is a function of
511
00:58:03,289 --> 00:58:11,469
temperature. And that is the resistivity of
the conductor changes with temperature. If
512
00:58:11,469 --> 00:58:17,259
the
temperature goes up resistivity will also
513
00:58:17,259 --> 00:58:23,660
increase. L is the length of the conductor
and A
514
00:58:23,660 --> 00:58:32,180
is the conductor cross sectional area. I would
like you to answer the following questions.
515
00:58:32,180 --> 00:58:33,180
.
516
00:58:33,180 --> 00:58:37,881
First is what type of conductors are used
for overhead transmission lines? Second
517
00:58:37,881 --> 00:58:45,839
question is how does conductor diameter affect
inductance of transmission line? And the
518
00:58:45,839 --> 00:58:54,130
.third question is why bundle conductors are
used in extra high voltage lines? So, I hope
519
00:58:54,130 --> 00:59:00,509
after going through this lesson, you should
be able to answer these questions.
520
00:59:00,509 --> 00:59:02,389
Thank you.
521
00:59:02,389 --> 00:59:02,389
.